I'm having issues getting my head around cartesian products and their cardinalities.
$A = \{0, 1, \{2, 3, 4\}\}$
$B = \{1,5\}$
$D = B \times N$ (where $N$ is the set of natural numbers)
The first problem: What is the cardinality of:
(a) $A \times B$ (cartesian product)
(b) $A \times D$
Part 2: true/false
(a) $N$ is a subset of $D$
for (a) I used $|A \times B|$ = $|A| * |B|$
and got $3*2 = 6$
is this the correct way to do this?
for (b) I assumed that the cardinality was infinite since it involved the set of natural numbers, am I correct in assuming this?
for part 2 (a) I assumed that it was true since $D$ contains the natural set so presumably the natural set is a subset of $D$, am I correct in assuming this?
Best Answer
$\mathbb N$ is not a subset of $D$. Subsets of $D$ alike to $\mathbb N$ are, for example $$\{1\} \times \mathbb N, \{5\} \times \mathbb N$$ The difference is that elements from $D$ look like $(1, a)$ or $(5, a)$ with $a\in\mathbb N$ whereas elements from $\mathbb N$ are just natural numbers (no tuples).
As for the cardinalities, you are right; $$|A\times B| = 6, |A\times D| = |D| = |\mathbb N| = \aleph_0 \quad \text{("countable infinity")}$$
More generally spoken, there are subsets of $A\times B$ looking like $A$ or $B$, namely sets of the form $A\times \{b\}, \{a\} \times B$ with $a\in A, b\in B$, but $A,B$ are no subsets of $A\times B$.